# A skiers jump and distance

1. Feb 5, 2009

### Jbright1406

1. The problem statement, all variables and given/known data
A skier leaves the ramp of a ski jump with a velocity of v = 8.0 m/s, θ = 15.0° above the horizontal, as in the figure. The slope is inclined at 50.0°, and air resistance is negligible. (Assume up and right are positive, and down and left are negative.)

(a) Find the distance from the ramp to where the jumper lands.

2. Relevant equations

3. The attempt at a solution

what im coming up with is
dcos50=(8cos15)t deltaX=Vxi(t)
-dsin50=(8sin15)t+.5(-9.8)t^2 deltaY=Vyt+.5(g)t^2

im stuck at how to solve for d and t. and i have to have this done by 11:30 :-) any helps on cracking this is more than appreciated. i think im supposed to solve for one of them and then plug it into the other equation but im not sure how to do that

2. Feb 5, 2009

### Nabeshin

Try solving for time in the y direction and substituting into the x equation to obtain a distance. This will involve solving a quadratic.

3. Feb 5, 2009

### Jbright1406

ok, my times up anyways but i still wanna figure it out. how do i solve for t in the y direction. cause in m equation i still have d right? i know the quadratic formule is x equals negative b plus or minus the sqr root of b^2-4ac all divided by 2a. i get lost there

4. Feb 5, 2009

### rl.bhat

Wright the projectile equation as

y = [tan(theta)]x - g*x^2/[(Vocos(theta)]^2
Substitute x = d*cos(theta) and d*sin(theta) and solve for d.

5. Feb 5, 2009

### Jbright1406

huh? i have alpha and beta. alpha is 15 beta is 50